The canonical crystal $B(\lambda )$ is a highest weight crystal of highest weight $\lambda$

The top residue. By [F1], $\mathcal{L}(\lambda {)}_{\lambda }$ is free of rank one with generator $v_{\lambda }$, so its residue space $\mathcal{L}(\lambda {)}_{\lambda }\mathrm{/}q\mathcal{L}(\lambda {)}_{\lambda }$ is one dimensional and $\mathcal{B}(\lambda {)}_{\lambda }=\{b_{\lambda }\}$ by [F3]; in particular $wt(b_{\lambda })=\lambda$. If ${\tilde{e}}_i{b}_{\lambda }\mathrm{\ne }0$ for some $i$, then by [F4] it is an element of $\mathcal{B}(\lambda {)}_{\lambda +{\alpha }_i}$ and is the residue of a nonzero homogeneous vector of $V(\lambda )$ of weight $\lambda +{\alpha }_i$; but $\mathrm{w}\mathrm{t}(V(\lambda ))\subseteq \lambda -Q_{+}$ by [F5], a contradiction since $\lambda +{\alpha }_i\mathrm{\notin }\lambda -Q_{+}$. Hence ${\tilde{e}}_i{b}_{\lambda }=0$ for all $i$, and $b_{\lambda }$ is a highest weight element of weight $\lambda$, proving (i).

Lowering-word generation. Clause (c) of [F2] states $\mathcal{B}(\lambda {)}_{\lambda -\alpha }=({\bigcup }_{i\in I}{\tilde{f}}_i\mathcal{B}(\lambda {)}_{\lambda -\alpha +{\alpha }_i})\setminus \{0\}$ for $\alpha \in Q_{+}\setminus \{0\}$. Induction on the height $\mathrm{ht}(\alpha )$ proves that every $b\in \mathcal{B}(\lambda {)}_{\lambda -\alpha }$ is an iterated lowering image ${\tilde{f}}_{i_1}\cdots {\tilde{f}}_{i_r}{v}_{\lambda }$ of $b_{\lambda }=v_{\lambda }+q\mathcal{L}(\lambda )$: for $\alpha =0$ this is step 1.1, and for $\alpha \mathrm{\ne }0$ the formula exhibits $b={\tilde{f}}_i{b}^{\mathrm{\prime }}$ with $b^{\mathrm{\prime }}\in \mathcal{B}(\lambda {)}_{\lambda -\alpha +{\alpha }_i}$ of strictly smaller height, to which the induction hypothesis applies. Hence $\mathcal{B}(\lambda )=F(b_{\lambda })$, proving (ii).

Killing by all ${\tilde{e}}_i$. Let $\mu \mathrm{\ne }\lambda$ and let $u\in \mathcal{L}(\lambda {)}_{\mu }\mathrm{/}q\mathcal{L}(\lambda {)}_{\mu }$ be nonzero with ${\tilde{e}}_{i}u=0$ for every $i$. Since $\mathcal{B}(\lambda {)}_{\mu }$ is a $\mathbb{Q}$-basis of this space by [F3] and [F1] shows it is finite (it has ${\dim }_{\mathbb{Q}(q)}V(\lambda {)}_{\mu }$ elements), $u={\sum }_{b\in \mathcal{B}(\lambda {)}_{\mu }}{a}_{b}b$ is a finite sum with coefficients $a_b\in \mathbb{Q}$. Fix $i$: the nonzero elements among ${\tilde{e}}_{i}b$, $b\in \mathcal{B}(\lambda {)}_{\mu }$, lie in $\mathcal{B}(\lambda {)}_{\mu +{\alpha }_i}$ and are pairwise distinct by the partial-inverse axiom [F3], hence linearly independent in the $\mathbb{Q}$-vector space $\mathcal{L}(\lambda {)}_{\mu +{\alpha }_i}\mathrm{/}q\mathcal{L}(\lambda {)}_{\mu +{\alpha }_i}$; the relation $0={\tilde{e}}_{i}u={\sum }_{b:{\tilde{e}}_{i}b\mathrm{\ne }0}{a}_b{\tilde{e}}_{i}b$ therefore forces $a_b=0$ for every $b$ with ${\tilde{e}}_{i}b\mathrm{\ne }0$. This holds for every $i\in I$. Now let $b\in \mathcal{B}(\lambda {)}_{\mu }$; by step 2.1 $b$ is a lowering word of positive length from $b_{\lambda }$ (since $\mu \mathrm{\ne }\lambda$), so $b={\tilde{f}}_i{b}^{\mathrm{\prime }}$ for the last letter $i$, and then ${\tilde{e}}_{i}b=b^{\mathrm{\prime }}\mathrm{\ne }0$ by [F3]; hence every $b$ occurs among the terms with ${\tilde{e}}_{i}b\mathrm{\ne }0$ for some $i$, so all $a_b=0$ and $u=0$, a contradiction. This proves (iii).

Conclusion. Steps 1.1, 2.1 and 3.1 establish (i), (ii) and (iii), so $B(\lambda )$ has a highest weight element $b_{\lambda }$ of weight $\lambda$ and $B(\lambda )=F(b_{\lambda })$: it is a highest weight crystal of highest weight $\lambda$ in the sense of [F7]. All arguments used finite sums and the fixed basis $\mathcal{B}(\lambda )$; no choice principle was used.